A) 46 kJ
B) 92 kJ
C) \[-\,92\text{ }kJ\]
D) \[-\,23\text{ }kJ\]
Correct Answer: B
Solution :
\[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)\xrightarrow[{}]{{}}N{{H}_{3}}(g);\] \[\Delta {{H}_{f}}=-46\,kJ\,mo{{l}^{-1}}\] \[{{N}_{2}}(g)+3{{H}_{2}}(g)\xrightarrow{{}}2N{{H}_{3}}(g);\] \[\Delta {{H}_{f}}=2\times -46=-92\,kJ\] \[2N{{H}_{3}}(g)\xrightarrow{{}}{{N}_{2}}(g)+3{{H}_{2}}(g);\] \[\Delta {{H}_{r}}=+92\text{ }kJ\] Hence,\[\Delta H\]of the reaction is +92 kJ.
You need to login to perform this action.
You will be redirected in
3 sec
